There is no special type for ($). The name is simply special cased in the
compiler. The rule is something like:
Whenever you see: f Prelude.$ x
instead try to type check: f x
That may not be the exact behavior, but it's close. To fix (.) (in a
similar fashion) you would have to have a similar rule, like:
Whenever you see: f Prelude.. g
instead try to type check: \x -> f (g x)
-- Dan
On Tue, Feb 10, 2015 at 6:19 PM, Tyson Whitehead <twhiteh...@gmail.com>
wrote:
> On February 10, 2015 16:28:56 Dan Doel wrote:
> > Impredicativity, with regard to type theories, generally refers to types
> > being able to quantify over the collection of types that they are then a
> > part of. So, the judgment:
> >
> > (forall (a :: *). a -> a) :: *
> >
> > is impredicative, because we have a type in * that quantifies over all
> > types in *, which includes itself. Impredicativity in general refers to
> > this sort of (mildly) self-referential definition.
>
> Thanks Dan and David,
>
> That was informative. Also very interesting that ($) is a special case.
> I tried this
>
> newtype Wrap = Wrap { extract :: forall f. Functor f => f Int }
>
> trip'' :: Wrap -> Wrap
> trip'' a = Wrap $ extract a
>
> and the compiler was happy. Wrapping ($) as ($') gave an error as you
> implied it would
>
> trip''' :: Wrap -> Wrap
> trip''' a = Wrap $' extract a
> where ($') = ($)
>
> With regard to my earlier comment about translating the (.) version
>
> trip' :: Wrap -> Wrap
> trip' = Wrap . extract
>
> to core, I can see it's actually okay. A most you may need is a lambda to
> float the implicit parameters backwards
>
> trip' :: Wrap -> Wrap
> trip' = Wrap . (\a f fDict -> extract f fDict a)
>
> as GHC seems to always float them as far leftward as possible
>
> extract :: Functor f => Wrap -> f Int
>
> I take it there are no user supplied types a person can give to overcome
> the predicative assumption?
>
> Out of curiosity, how would you write the special internal type that ($)
> has that separates it from ($') above?
>
> Thanks! -Tyson
>
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